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Mat he m

at ico ru m

CRUX MATHEMAT1C0RUM

Volume 16 §2

February I Fevrier

1990

CONTENTS / TABLE DESMATIERES

The Olympiad Comer: No. 112

Mini-Reviews

Problems 1511-1520

R.E. Woodrow

Andy Liu

88

U2

AS

Solutions 1898-U02

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Canadian Mathematical Society I Societe mathematique du Canada

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- 33 -

THE OLYMPIAD CORNER No. 112

R.E. WOODROW

All communications about this column should be sent to Professor R.E. Woodrow, Department of Mathematics and Statistics, The University of Calgary, Calgary, Alberta, Canada, T2N IN4.

First a correction and apology. The problems for the 24th Spanish Olympiad which we presented in the December 1989 number of the Corner were forwarded to us from Willie Yong of Singapore. Willie Yong also sent us the following problems from the XIV "All Union" Mathematical Olympiad of the U.S.S.R. The translation was done by Mark Saul. We give ten problems this month and the remaining ten next month.

XIV "ALL UNION" MATHEMATICAL OLYMPIAD (U.S.S.R.)

1. All the two-digit numbers from 19 to 80 are written in a row. The result is read as a single integer 19202122...787980. Is this integer

divisible by 1980?

Z. Side AB of a square A BCD is divided into u segment s in such a way that the sum of lengths of the even- A

numbered segments equals the sum of the lengths of 1

the odd-numbered segments. Lines parallel to AD 2

are drawn through each point of division, and each 4

of the n "stripsSf thus formed is divided by diagonal BD into a left region and a right region. Show that the sum of the areas of the left regions with a

odd numbers is equal to the sum of the areas of the n

right regions with even numbers. B

3. A, payload, packed into containers3 is to be delivered to the orbiting space station "Salyut". There are at least 35 containers, and the total

payload weighs exactly 18 tons. Seven "Progress" transport ships are available, each of which can deliver a 3-ton load. It is known that these ships altogether can (at least) carry any 35 of the containers at once. Show that in fact they can carry the entire load at once.

- 34 -

4. Points M and P are the midpoints of sides BC and CD of convex quadrilateral ABCD. If AM + AP = a, show that the area of the

region ABCD is less than a2/2.

5. Does the equation x2 + yz = z4 have solutions for prime numbers x9 y and z?

6. Given point E on diameter AC of a circle, construct chord BD passing through E and such that the quadrilateral ABCD has the largest

possible area.

7. Several points are located along the shore of a circular lake. Some of these are linked by ferry service. Points A and B are linked by a

ferry if and only if the next two adjacent points A' and B' (proceeding clockwise around the lake) are not linked. Show that one can go by ferry from any point to any other point, changing boats not more than twice.

8. A number is written (in base 10 notation) using six distinct non-zero

digits, and it is divisible by 37. Show that by permuting the digits one can obtain at least 23 different new numbers, each divisible by 37.

9. Solve simultaneously: sin x + 2 sin(x + y + z) = 0 sin y + 3 sin(z + y + z) = 0 sin z + 4 sin(x + y + z) = 0.

10. A set of 1980 vectors are given in the plane, not all of which are collinear. It is known that the sum of any 1979 of these vectors is

collinear with the one vector not included in the sum. Prove that the sum of all 1980 given vectors is equal to the vector 0.

* * *

As I mentioned in last month's number, it is now time to begin giving the results to some of the challenges to solve remaining problems. The response has been very good. The first of these are to problems from Volume 7 (1981). Several of these will be attributed to Anonymous. A reader sent in a package of solutions but did not indicate his/her name on the sheets. Unfortunately, the envelope has disappeared and short of calling in handwriting experts I don't see how to properly attribute these solutions. I would be delighted if Anonymous would write to reestablish contact!

- 35 -

1 [1981: 15] 86th Moscow Olympiad (1973).

A square island consists of several estates. Can one divide these estates into smaller estates in such a way that no new intersection points of boundaries are introduced, and so that the entire map of the island can be coloured with only two colours (estates with a common boundary being coloured differently)?

Solution by Anonymous.

In graph theoretic language, we have a planar map where the estates are the faces, the boundaries are the edges, and the intersection points of boundaries are the vertices. The answer depends on whether multiple edges are permitted. If not, no further subdivision of the map in Figure 1 is possible, and this map is clearly not two-colourable. If multiple edges are allowed, the answer is trivially affirmative. One can simply double up every inland edge. Colour all the crescent-shaped new faces created in one colour, and the remaining faces in the other colour. The result of applying this process to Figure 1 is Figure 2.

Figure 1 Figure 2

8. [1981: 15] 36th Moscow Olympiad (1973). The faces of a cube are numbered 1,2,...,6 in such a way that the sum

of the numbers on opposite faces is always 7. We have a chessboard of 50 * 50 squares, each square congruent to a face of the cube. The cube "rolls" from the lower left-hand corner of the chessboard to the upper right-hand corner. The "rolling" of the cube consists of a rotation about one of its edges so that one face rests on a square of the chessboard. The cube may roll upward and to the right (never downward or to the left). On each square of the chessboard that was occupied during the trip is written the number of the face of the cube that rested there. Find the largest and the smallest sum that these numbers may have.

Solution by George Evagelopoulos, Law student, Athens, Greece, and by

Anonymous.

The mean value of the sum of the numbers written on the chessboard is equal to the number 3.5 * 99 = 346.5. If during the rolling of the cube we meet twice

- 36 -

the number a, then between these two times we will also meet the number 7 - a. Thus for each a the numbers of times we will meet the numbers a and 7 - a will either be equal or will differ by 1. Consequently, the sum of the written numbers can differ from the mean value of the sum by at most 0.5 + 1.5 + 2.5 = 4.5 so the largest value for the sum is 351 and the smallest is 342.

9. [1981: 16] 36th Moscow Olympiad (1973). On a piece of paper is an inkblot. For each point of the inkblot, we

find the greatest and smallest distances from that point to the boundary of the ink­blot. Of all the smallest distances we choose the maximum and of all the greatest distances we choose the minimum. If these two chosen numbers are equal, what shape can the inkblot have?

Solution by Anonymous and by George Bvagelopoulos, Law student, Athens,

Greece.

Let A be a point such that its greatest distance from the boundary of the inkblot is r and that r is minimum among all points. Then the inkblot is enclosed by the circle 7 with center A and radius r. Let B be a point such that its smallest distance from the boundary of the inkblot is r where r is maximum among all points, by hypothesis. Suppose A and B are distinct. Then there is a boundary point C of the inkblot such that B lies on the segment AC. Now we have BC > r and AC < r, a contradiction. Hence A and B coincide. If there were a boundary point D inside of 7, then BD < r is a contradiction. Hence the inkblot is the disc enclosed by 7.

13. [1981: 16] 36th Moscow Olympiad (1973). The following operation is performed on a 100 digit number: a block of

10 consecutively-placed digits is chosen and the first five are interchanged with the last five (the 1st with the 6th, the 2nd with the 7th,..., the 5th with the 10th). Two 100—digit numbers which are obtained from each other by repeatedly performing this operation will be called similar. What is the largest number of 100—digit integers, each consisting of the digits 1 and 2, which can be chosen so that no two of the integers will be similar?

Solution by Anonymous.

A digit in the nth position is said to belong to the ith orbit, 0 < i < 4, if n = i mod 5. In a 100 digit number, each orbit consists of 20 digits. For numbers consisting of l's and 2's only, two numbers are said to be alike if they have the same number of l's in each pair of corresponding orbits. Since there are 21 possible

- 37 -

numbers (from 0 to 20) of l?s in each orbit, the maximum number of numbers no two of which are alike is 215. Since the operation allowed does not take any digit out of its orbit, two numbers that are similar must be alike. We claim that, conversely, two numbers that are alike are also similar. Our basic transformation consists of the following sequence of operations:

a\ 0,2 03 04 a§ a% at a% a% a\o an ... First

fl6 07 08 09 010 01 02 0 | 04 0§ 011 •••

Second ©6 02 03 04 05 011 07 08 09 010 01 •-

Notice this permutes the digits in one orbit only, which allows us to fix up the number one orbit at a time. Notice, too, that in each orbit three consecutive digits may be permuted cyclically. Consider a given arrangement &i,&29—>&2o of lJs and 2's in an orbit and a permutation &1<7,&20')--->&2O0' of them. Either this permutation or bia,b2(T,..-Ms(TMQ(r>bi90r is even and so is produced via successive applications of the basic operations. Now if b\$a and 620 are both l3s or both 2Js we are done. Thus we may suppose that 619a = 1 and 620 = 2, while we have produced the reversed arrangement. If b\%a = 1 the desired orbit ends 112 while we produce 121. If b\%a = 2 the desired orbit ends 212 while we see 221. In either case a cyclic permutation of these three positions will produce the desired arrangement. It follows that the maximum number of pairwise dissimilar numbers is 215.

15. [1981: 16] 36th Moscow Olympiad (1973).

On each endpoint of a line segment the number 1 is placed. As a first step, between these two numbers is placed their sum, 2. For each successive step, between every two adjacent numbers we place their sum. (After the second step we have 1, 3, 2, 3, 1; after the third we have 1, 4, 3, 5, 2, -5, 3, 4, 1; and so on.) How many times will the number 1973 appear after a million steps?

Solution by R.K. Guy, The University of Calgary, and by Anonymous. 1. Consecutive numbers after any step are prime to one another, because

this is true for the first few steps, and gcd (a,b) = 1 implies gcd (a,a + b) = 1 = gcd (a + 6,6).

2. Every pair of coprime integers will appear at some stage, for (a,b)

appears after (a - 6,6) or (a,b - a) has appeared, according as a > 6 or a < 6, and

these last will have appeared by induction.

3. 1973 appears in the step after (a,6) where a + 6 = 1973 and

gcd (a,6) = 1, i.e. whenever 1 < a < 1973, and gcd (a,1973 - a) = 1, that is <p(1973)

times, where <p is Euler's totient function, the number of numbers not exceeding 1973

and prime to it. As 1973 is prime, ^(1973) = 1972, which is the number of times

- 38 - .

1973 occurs, provided there have been at least 1972 steps, which is less than one million.

4. Notice that the numbers will all eventually occur, and (as a subsequence) in the same order, even if we don't insert every possible number at each step. For example, if at step n we only insert n + 1 (and no larger sums):

1 1 1 2 1

1 3 2 3 1 1 4 3 2 3 4 1

1 5 4 3 5 2 5 3 4 5 1

Then all the occurrences, ip(n + 1) of them, of n + 1 occur at the same step, and

the sequence is that of the denominators of the Farey series of order n + 1 [Hardy and Wright, Introduction to Theory of Numbers, 4th Edition, Oxford Univ. Press, 1959? Ch. 3], which has many interesting properties.

* * *

Now we return to solutions to problems from the April 1988 number of the Corner* Some readers sent in solutions to problems from the AIMEy which we shall not reproduce here because of the solutions booklet available from the M.A.A. But thank you for sending them along. The solutions we do give are to problems of the Tenth Undergraduate Mathematics Competition of the Atlantic Provinces Council on the Sciences.

JL [1988: 101] Tenth Atlantic Provinces Mathematics Competition.

Find all different right-angled triangles with all sides of integral length

whose areas equal their perimeters.

Solution by the late J.T Groenman, Amhem; The Netherlands. Let the two legs have lengths x and y and the hypotenuse be z. Then

x2 + y2 = z2 and | ^ = x + y + z .

This gives / \ 2 2 2

%2 + y2 = (2F " x " y\ = 4 * + %2 + y2 ~ **v ~ %y2 + 2xy -From this

xy(xy - Ax — Ay + 8) = x*y2 - Ao?y - ixy2 + Sxy = 0 . Since we rule out degenerate triangles, xy - Ax - 4y + 8 = 0 and so

- 39 -

(x - 4)(y - 4) = 8 . •.

Supposing x > j / , we have either x - 4 = 8, 3/ - 4 = 1 and so x = .12, y = 5, z = 13, o r x - 4 = 4, y - 4 = 2 and so x = 8, y = 6, z = 10. Therefore the only triangles with the given properties are the 5, 12, 13 and 6, 8, 10 right triangles.

2. [1988: 101] Tenth Atlantic Provinces Mathematics Competition.

Sketch the graph and find the measure of the area bounded by the relation

I s - 6 0 ] + |y| = )x/4[ .

Solution by the late J.T. Groenman, Amhem% The Netherlands. (60,15) Note

|x/4| - | i - 601 = If x > 60 we have

| - (x - 60) = 60 -

\v\ > o

- | * > o giving x < 80 and y = ±(60 - 3i/4). If 0 < x < 60 we have

| - (60 - x) = | i - 60 > 0,

and x > 48, y = ±(5i/4 - 60). Finally if x < 0 we have

- | - (60 - x) = | , - 60 > 0 ^ 6 0 ) _ 1 5 )

giving x > 80, which is impossible. The area is (30°32)/2 = 480.

4. [1988: 102] Tenth Atlantic Provinces Mathematics Competition.

Given a function / and a constant c # 0 such that (i) f(x) is even, (ii) g(x) = /(x - c) is odd,

prove that / is periodic and determine its period.

Solution by Bob Prielipp, University of Wisconsin-Oshkosh.

Let x be an arbitrary element of the domain of / . Then f(x - 4c) = g(x - 3c) = -g(3c - x) [because g(x) is odd]

= -/(2c - x) = -f(x - 2c) [since f(x) is even] = -$(& - <0 = $(c - x) = /(-a?) = /(&).

It follows that / i s periodic with a period of 4 |c | . The prescribed conditions do not guarantee that / has a smallest positive

period. When / is the zero constant function, g is also the zero constant function. The zero constant function is both odd and even. Each positive real is a period, so

- 40 -

there is no smallest positive period.

[Editor's note: this partially answers Crux 1497 [1989: 298].]

5. [1988: 102] Tenth Atlantic Provinces Mathematics Competition.

Show that

^ < exp J ^L£ dx 1 < 3 .

Solution by Robert E. Shafer, Berkeley, California.

Note TT/2 _TT/2

'x/6 x

Now f(t) = (sin £)/£ has

/sin x , f 4 „ z „ z . x .

—— dx = J - cos r COST smT ax .

/ ' ( * ) =

V/6 * *

t cos f - sin f

f2

and so / ' ( £ ) < 0 on (0,7r/2], since g(t) = tan t - t has #'(£) = sec2£ - 1 > 0 giving

a minimum of 0 at t = 0. Thus sin(x/4)

x/4

is decreasing on [7r/6,7r/2] giving that

s i * * { 2 4 ) J cos* cos? dx and V/6

sin .w/2

'TT/6

in(7r/8) f „^„x „„„x . >g' / J COSTT C O S T OX

are respectively an upper bound and a lower bound to

,TT/2

JTT/6 X

But .x/2

'x/6 J cos£ cos? da;

,TT/2

| s in |^ + js inj - 2 sin*"

I COST + COS-?|rfx

2T

Thus, the lower bound is

_5

6y^

while the upper bound is

L = if2 _ JL + & _ #_] .=• s T 6v ^ 2yd ~ *

1 _ _l

IV2 v J 1 - - 1 + -1

V? VSJ i + JZ + fi

&

*-?[ - 5 + -1 _ -1 + M V2" yfi 3 ^ .

24 7T

r \ /

— - —

ys v^Jl 11--! + -*

V2" V5J (Not too pretty, perhaps, but reasonably precise, as the estimates are within 3% of

- 41 -

each other.) Now L > 0.84 and e0'84 > 2.3 > # while U < 0.87 and e0'87 < 2.4 < 3. This gives the result.

6. [1988: 102] Tenth Atlantic Provinces Mathematics Competition. Three equal circles each pass through the centres of the other two.

What is the area of their common intersection?

Solution by the late J.T. Groenman^ Amhem$ The Netherlands.

Let the centres of the circles be Oh

02, 0$ and let the radius be r. Let A be the area of the triangle O1O2O3 and let 5 be the area of the sector of the circle with centre 0\ subtended by chord 020%. Then the required area is

A + 3(5 - A) = 35 -2A

= 3.J*r* - 2 * J f ^

7, [1988: 102] Tenth Atlantic Provinces Mathematics Competition. Prove that

(19872)! (1987!)1988

is an integer.

Comment by Bob Prielipp, University of Wisconsin-Oshkosh.

The following more general result is known: (ab)l

al{bl)a

is an integer where a and 6 are non-negative integers. [For a proof of this see p. 103 of Niven and Zuckerman, An Introduction to the Theory of Numbers (4th Edition).]

* * *

This exhausts the solutions sent in to problems from the April 1988 column. Don't forget to send me the contests from around the world as well as your

solutions to problems posed! * * . *

- 42 -

M I N I - R E V I E W S by

ANDY LIU

OXFORD UNIVERSITY PRESS SERIES ON RECREATIONS IN MATHEMATICS

This hardcover series is edited by David Singmaster of the Polytechnic of the South Bank in London. He is well-known for his writing on recreational mathematics, and is the leading expert on the history of the subject. The four volumes published so far are new, but the series may contain translations and reprints of classic works.

Volume 1. Mathematical Byways in Ayling, Beeling and Ceiling, H. ApSimon, 1984. (97 pp.) This book contains eleven chapters, each built around a central problem, with

solutions and generalizations. Each problem is in the form of an event in the three villages named in the title, though there are occasional wanderings off to Dealing and beyond. The titles of the problems are ladder-box, meta-ladder-box, complete quadrilateral, bowling (as in cricket) averages, centre-point, counting sheep, transport, alley ladder, counterfeit coins, wrapping a parcel and sheep-dog trials.

Volume 2. The Ins and Outs of Peg Solitaire, J.D. Beasley, 1985. (275 pp.) This book deals mainly with peg solitaire on the standard "English" board

with 33 holes, each of which can hold one man. Each move consists of a jump by one man over an adjacent one into an empty hole, or a continuous sequence of such jumps. The normal objective is to convert a starting configuration into a target position with fewer men (as those jumped over are removed). The book presents a balanced treatment of the mathematical ideas behind the puzzle as well as the actual techniques in solving it.

Volume 3. Rubifcs Cube Compendium, E. Rubik, T. Varga, G. Keri, G. Marx and T. Vekerdy, 1986. (225 pp.) This book consists of six articles, and an informative update by David

Singmaster. The lead article is written by the inventor of this mathematical phenomenon, giving some insight into how the idea was originally conceived. The other authors are Rubik's compatriots, and their articles deal with the mathematics and the techniques of "cubing". The book has many brightly coloured illustrations.

- 43 -

Volume 4. Sliding-Piece Pwszks, L.B.. Hordem, 1986. (249 pp.) The classic example of a sliding-piece puzzle is the 14-15 puzzle of Sam Loyd.

This and 271 other puzzles are catalogued by class in this book. In almost all cases, there is enough information for home-made copies to be constructed. There is also a colourful history of the subject.

* * *

P R O B L E M S Problem proposals and solutions should be sent to the editor, B. Sands,

Department of Mathematics and Statistics, University of Calgary, Calgary, Alberta, Canada, T2N 1N4- Proposals should, whenever possible, be accompanied by a solution, references, and other insights which are likely to be of help to the editor* An asterisk (*) after a number indicates a problem submitted without a solution.

Original problems are particularly sought But other interesting problems may abo be acceptable provided they are not too well known and references are given as to their provenance. Ordinarily, if the originator of a problem can be located, it should not be submitted by somebody else without his or her permission.

To facilitate their consideration, your solutions, typewritten or neatly handwritten on signed, separate sheets, should preferably be mailed to the editor before September 1, 1990, although solutions received after that date will abo be considered until the time when a solution is published.

1511. Proposed by J.T. Groenman, Arnhem, The Netherlands.

Evaluate n

l im TT(1 - t a n 4 £-).

1512. Proposed by Walther Janous, Ursulinengymmsium; Innsbruck^ Austria.

Given r > 0, determine a constant C = C(r) such that

(1 + Z)T(1 + ZT) < C(l + 2?)T

for all z > 0.

1513. Proposed by M.S. Klamkin, University of Alberta. (a) A planar centrosymmetric polygon is inscribed in a strictly

convex planar centrosymmetric region R. Prove that the two centers coincide. (b) Do part (a) if the polygon is circumscribed about R.

(c) Do (a) and (b) still hold if the polygon and region are n-dimensional for n > 2?

- 44 -

1514. Proposed by George Tsintsifas, Thessaloniki, Greece. Let ABC be a triangle with sides a, 6, c and let P be a point in the

same plane. Put AP = Ru BP = R2, CP = iE3. It is well known that there is a triangle with sides aRh bR2, cR$. Find the locus of P so that the area of this triangle is a given constant.

1515. Proposed by Martin E. Kuczma, Warszawa, Poland. We are given a finite collection of segments in the plane, of total

length 1. Prove that there exists a straight line I such that the sum of lengths of projections of the given segments to line I is less than 2/w.

1516. Proposed by Toshio Seimiya, Kawasaki, Japan. ABC is an isosceles triangle in which AB -AC and zA < 90°. Let

D be any point on segment BC. Draw CE parallel to AD meeting AB produced in E. Prove that CE > 2 CD.

1517. Proposed by Bill Sands, University of Calgary. [This problem came up in a combinatorics course, and is quite likely

already known. What is wanted is a nice answer with a nice proof. A reference would also be welcome.]

Imagine you are standing at a point on the edge of a half-plane extending infinitely far north, east, and west (say, on the Canada-USA border near Estevan, Saskatchewan). How many walks of n steps can you make, if each step is 1 metre either north, east, west, or south, and you never step off the half-plane? For example, there are three such walks of length 1 and ten of length 2.

1518. Proposed by K.R.S. Sastry, Addis Ababa, Ethiopia. O and H are respectively the circumcenter and orthocenter of a

triangle ABC in which zA # 90°. Characterize triangles ABC for which AAOH is isosceles. Which of these triangles ABC have integer sides?

1519. Proposed by P. Penning, Delft, The Netherlands. Find all prime numbers which, written in the number system with

base 6, contain each digit 0,l,---,6 - 1 exactly once (a leading zero is allowed).

1520. Proposed by Stanley Rabinowitz, Westford, Massachusetts. A point is said to be inside a parabola if it is on the same side of the

parabola as the focus. Given a finite number of parabolas in the plane, must there

be some point of the plane that is not inside any of the parabolas? * * *

- 4 5 -

S O L U T I O N S No problem is ever permanently closed. The editor will always be pleased to

consider for publication new solutions or new insights on past problems.

1393. [1988: 302] Proposed by XT. Groenman, Arnhem, The Netherlands.

Let AtA2A$ be a triangle with incenter 7, excenters 7i, 72? h, and median point G. Let Ht be the orthocenter of AItA2A$3 and define H2} Hz

analogously. Prove that AiH^ i.2^21 A$Hz are concurrent at a point collinear with G and I

Solution by P. Penning^ Delfts The Netherlands. We note that ^3 is located on the straight line 7i72, with A* the pedal point

of /, since both ^ s and 7^3 are perpendicular to 743 (outer bisectrix perpendicular to inner bisectrix). So AA1A2A2 is the pedal triangle of 7 in A7i727s. In fact, for any triangle 7^73, an arbitrary point 7 inside this triangle^ and A1A2AZ its pedal triangle, the property given in this problem holds.

Define as origin 7, and the vectors Ai = JTl , etc.

Now

Hi = A2 + A3 , H2 = A3 + At , Hi = Ai + A2 ,

because the quadrangles IA2HtA^ etc. are all three parallelograms. Since

Ai + Hi _ At + A2 + A3 p t r

it is now easy to see that the point S such that

q _ Ai + A2 + A3 l\

is the midpoint of all three straight lines A\H\, A2H2, AzH$. Since the median point G is located at (At + A2 + A3)/3 it follows that S lies on the line through J and G with

7G = 2US .

Also solved by FRANCISCO BELLOT ROSADO, LB. Emilio Ferrari, and

MARIA ASCENSION LOPEZ CHAMORRO, LB. Leopoldo Cano, Valladolid, Spain;

L.J. HUT, Groningen, The Netherlands; T. SEIMIYA, Kawasaki, Japan; D.J.

SMEENK, Zalthommel, The Netherlands; DAN SOKOLOWSKY, Williamsburg,

Virginia; JOSE YUSTY PITA, Madrid, Spain; and the proposer.

The above generalization was also proved by Hut.

- 46 -

1394. [1988: 302] Proposed by Murray S. Klamkin, University of Alberta.

If x, y, z > 0, prove that

Jy2 + yz+z2 + Jz* + zx+x2 + Jx*+xy+y2> 3/yz + zx + xy .

I. Solution by Jack Garjunkel, Flushing, New York Note that

<$(z + V + z) I $JxyT+ yz + zx holds, since it is known that

(x + y + z)2 > 3(xy + yz + zx) .

Thus we will prove the stronger inequality

/^+yz+^ + fz2 + zx + x* + J~x2 + xy +y2 > # ( z + y + z) . (1) Now

J y2 + yz+ z2 >^y + z) (2)

holds, since it is equivalent to iy2 + 4yz + iz2 > 3y2 + 6yz + Zz2

or

(y - z)2 > o . Similarly

J z> + zx + x2 > *§(* + x) and J x2 + xy + y2 > (x + y) ,

which with (2) proves (1).

IL Solution by Sa'ar Hersonsky, student, Technion, Haifa, Israel

We show that TT (x2 + xy + y2) > (xy + yz + zxf , (3)

from which the given inequality follows by the AM-GM inequality:

( \ ]T 7 x2 + xy + y2 1 > TT 7 x2 + xy + y2 > (Jxy + yz + zxf .

(The above sum and products are cyclic over x, y, z.) When multiplied out, (3)

becomes Y x4yz + Y xAy2 + 2 Y xzy2z + Y xzyz + $x*y2z2

- X ^yZ + 3 X x 3 ^ + 6rc2^2

(each sum over all symmetric versions of its summand), or

Y z V + y\ *V l y\ 3?y2* + %z*y2z2 •

- 47 -

By the AM-GM inequality,

£ x4yz > 3 fx*y«? = 3SY*2 ,

\ x*y2 > \ 3?y2z . so it is enough to prove

But this is just

which is obvious.

[Editofs note. Upon being told of Bersonskyfs solution Murray Klamkin offered the following very short proof of inequality (3), using Holder's inequality:

zy+yz+ zx= W ' W ' V ) 1 ' 8 + W ' W ' W ' 1 + ( * T ' W ' W ' S

< (xy + y2 + ^)u\y2 + yz + #)ul{?? + z2 + JOT)1'3

- J[ (x2 + xy + y2)^ , and (3) follows.]

III. Solution hy Walther Janous, Ursulinengymnasium^ Innsbruck, Austria. We show more generally: if a, '/?, 7 > 0 sac/i that a + / ? + 7 = 2 7 r ?

E 7 x2 - 2xy cos 7 + ^2 £

Let P be a point of the plane and consider the triangle ABC generated by x, y,

z, a, P, 7 as indicated by the figure. Then

AB = v i2 - 2xy cos 7 + y2 , etc.

By item 4.2 of [1],

5 > J ZJ5F (5) where s is the semiperimeter and F the area of A ABC. Since

1

(4)

F = Jkmm

xy sin 7

inequality (5) reduces to (4). The case a = 0 = 7 = 2T/3 yields the stated result. It should be noted that any triangle inequality /(o,6,c,F,...) > 0 can be

transformed into an algebraic inequality i(sc,y,2,a,/?,7) > 0 via the x - y - z -

a - p _ 7 transformation as indicated above (i.e.

a = 7 1/2 - 2j/z cos a + z2 , etc.)

Especially,

- 48 -

2= fl, y = b , z = c , a = f - 4 , 0 = w - B , 7 = TT - (7 leads to the new triangle with sides

J b2 - 26c COS(TT - 4) + c2 = / 26^ + 2c2 - a2 = 2ma , etc.

where ma, ^b* ^c are the medians of AABC. Thus- the above defined class of transformations contains as a special case the median-duality (cf. [2], pp. 109-112).

We now give three applications of the transformation in case x, y, z > Q and a=7r-A,/3=w-Bij=w-C.

(i) The known inequality

a2 + b2 + c2 > 4 F # ([1], item 4.4) becomes

\ x2 + V yz cos A > J5 V |/z sin A ,

i.e.

I 1 2 > * X * H* - J) • (ii) Putting x = y = z in (4) we get

Y V 1 + cos A > J 3 # Y sin 4 ,

i.e.

E Ma A s 3/5^ c o s 7 ^ J T I T »

where s is the semiperimeter and R the circumradius of AABC.

(iii) Starting from the Neuberg-Pedoe two-triangle inequality

Y a'2(-a2 + b2 + e2) > 16FF'

([1], item 10.8), we get upon setting

a'2 = y2 + 2yz cos >4 + z2 , etc.

that

Y (y2 + z2 + 2j/z cos A)bc cos i4 > 4F \ yz sin A ,

or, since 2F = be sin J4, etc.,

Y x2a(b cos C + c cos 5) + 2 Y ya*c cos2^4 > 2 Y y^6c sin2^4 .

Collecting terms and noting b cos C + c cos B = a and sin2i4 - COS2J4 = -cos 2 4 we

get

Y a2x2 > -2 Y 2/ 6c cos 2 4 ,

or equivalently (via a = 2R sin 4, etc.)

- 49 -

V 3?sin2A > -2 V yz sin B sin C cos 2 A .

References:

[1] 0. Bottema et al, Geometric Inequalities., Groningen, 1968. [2] B.S. Mitrinovic, J.E. Pecaric and V, Volenec, Recent Advances in Geometric

Inequalities, Dordrecht, 1989.

Also solved by SVETOSIAV J. BILCHEV and EMILIA A. VELIKOVA,

Technical University, Russe, Bulgaria; KEE-WAI LAU, Hong Kong; and the proposer.

Bilchev and Velikova, and the proposer, used a method similar to solution III.

1395, [1988: 302] Proposed by Walther Janous, Ursulinengymnasium,

Innsbruck, Austria. Given an equilateral triangle ABC, find all points P in the same plane

such that (PA)2, (PB)2, (PC)2 form a triangle.

Solution by Murray 5. Klamkin, University of Alberta. Let the vertices of the equilateral triangle be given by the points 1, <j and UJ2

in the complex plane, where UJ (f 1) is a cube root of unity. It follows by Ptolemy's inequality that PA, PB, PC are sides of a triangle (possibly degenerate) for all P in the plane or out of the plane [e.g. see Crux 1214 (1988: 118)]. The conditions in the current problem require that PA, PB, PC are sides of a non-obtuse

triangle (possibly degenerate, e.g. if P = A, B or C). If P lies in the domain of zBOC (0 the origin), then PA > PB and

PA > PC, and it suffices to require-(PB)2 + (PC)2 > (PA)2. If P has the complex representation z - x + iy, then this condition becomes

| z - UJ\2 + \ z - UJ2\2 > | z - 1|2 ,

or (x-cos 120)2 + (y-sin 120)2 + (x-cos 240)2 + (y-*in 240)2 > (x-4)2 + y2 .

The latter reduces to (x + 2)2 + y2 > 3 .

Thus, for this case, P must lie on or outside a circle of center (-2,0) and radius <J5. This circle is tangent to lines OB and OC at B

and C respectively. By symmetry, P must also lie on or outside the two circles obtained by rotating the above circle 120 and 240 about O as shown in the figure.

- 50 -

If we allow P to lie outside the plane it follows easily that P must lie on or outside three spheres whose centers and radii are the same as for the three previous circles. Also, the triangle formed is degenerate only when P lies on the boundary of the three spheres.

For a general triangle ABC of sides a, b, c and any point P in or out of the plane of ABC, it again follows from Ptolemy ?s inequality that a* PA, b*PB, c°PC are sides of a triangle (possibly degenerate). In order for the latter triangle to be non-obtuse, P will have to lie on or outside three spheres if ABC is acute; on or outside two spheres and on or on one side of a plane if ABC is right-angled; and on or outside two spheres and inside one sphere if ABC is obtuse. To see this let the vertices of ABC be given by A = (0,0), B = (c,0), and C = (u,v) where

u2 + v2 = b2 and (u - c)2 +• v2 = a2 . If P = (x,y,z), then

(PA)2 = x2 + y2 + z2 ,

(PB)2 = (x - c)2 + y2 + z2 , (PC)2 = ( x - ti)2 + ( y - t/)2 + z2 ,

and our conditions become a2(x2 + y2 + z2) + b2[(x - c)2 + y2 + z2] -~ c2[(x - u)2 + (y - t/)2 + z2] > 0 , a\x2 + y2 + z2) - b2[(x - c)2 + y2 + z2] + c2[(x - u)2 + (v - *)2 + *2] > 0 ,

-a2(x2 + y2 + z2) + 62[(x - c)2 + y2 + *2] + c2[(s - u)2 + (y - t/)2 + *2] > 0 . The above result follows.

The given problem can be generalized another way by considering an n-dimensional regular simplex whose vertices AQ,A\,- • *,An are the endpoints of n + 1 concurrent unit vectors A0,Ai,---,AR and determining all points P such that {PAQ)2,

(PAI)2, ---, (PAn)2 are sides of an (n + l)-gon. We claim that if n > 3, any point

P will do. Here we must satisfy

(P - A0)2 + (P - A|)2 + . . . + (P - Aa)

2 > 2(P - A*)2

for k = 0,1,...,n. Squaring out and using the fact that \ &* ~ ®, w e obtain the

following T i + 1 conditions on P: (TI - 1)(P2 + 1) + 4P-A& > 0 , k = 0,1,...,TI . (1)

The case for n - 2 reduces to the given problem which was solved above. For n = 3 + m (m > 0), (1) becomes

m(P2 + 1) + 2(P + A*)2 > 0 , and this is clearly satisfied for all P. Furthermore, since

- 51 -

X ip-Afci { k - 1

11.

> ]T I P - A f c [ 2 > |P - Ao|2 , etc.,

fc*i

it follows that Pj4o,Pj4i,...,PAn are also sides of an (n + l)-gon for all n > 2 and for all P. The case for n = 2 was taken care of previously, using Ptolemy's inequality. More generally, let F(x) be any nonnegative, nondecreasing concave function with F[0) = 0. Then

n |

X n i P - A * | 2 ) > F\ k= l

IP - A, | 2

1 **i

> *(|P - A0f2) , etc.

As an example, we can take F(x) = xr for 0 < r < 1. It then follows that | P - A, | 2 - , * = 0fl,...,n

are sides of an (n + l)-gon.

-Afao 5ofosi by JORDI DOU, Barcelona, Spain; HANS ENGELHAUPT,

Gundelsheim, Federal Republic of Germany; RICHARD I. HESS, Rancho Palos

Verdes, California; P. PENNING, Delft, The Netherlands; and the proposer.

The extension of the problem to an equilateral triangle ABC and a point P in

3 -space (or even n-space) was also done by the proposer.

1398. [1988: 302] Proposed by Colin Springer, student, Waterloo, Ontario.

Evaluate

where n is a positive integer, n > 2.

L Solution by C. Wildhagen, Breda, The Netherlands.

We begin with the obvious identity

n - l

^ - 1 = (z - 1) TT (z - a2***'*) .

Dividing both sides by z - 1 and putting z = 1 gives

n - l

n = TT (1 - e2*11'"1) . k » l

Take the complex conjugate of both sides of (1) to obtain

(i)

- 52 -

n - i

n = TT (1 - a"2***'*) . (2)

Multiplying (1) and (2) yields

n - l n - l

n2 = TT (2 - #***'* - e-afcui^) = 2*"1 TT (l - cos^p] ,

fc«l fe*l

hence n - I

II. Solution by Murray S. Klamkin, University of Alberta. This is a special case of a well known result given in good

trigonometry books (see [1]). "The equation x2n - 2a^cos not + 1 = 0

is a quadratic in af1, with roots cos na ± i sin na; thus the 2n values of x are ^ . L i 2ibr| . . f , 2fori

for fc = 0,1 ,...,n - 1." This leads to the factorization

n - l

x2*1 - 2xRcos na + 1 = TT fx2 - 2x cosfa + ~ [ ] + l] .

Many results can be obtained from this identity by specializing x and a. In particular, let a = 0 to give

( x - 1 ) 2 1 1 ^ n >

Now let x -» 1 to give

„> . 2 - J ] (l - cos^) = 2 - J ] *•(£) . fc = l Js = 1

Reference: [1] C.V. Durell, A. Robson, Advanced Trigonometry, G. Bell, London, 1953,

p. 227.

III. Solution by Stanley Rabinowitzr Westford, Massachusetts.

For n > 2, let

- 53 -

P-flf!-*^) k » l

Applying the substitution 1 - cos 2x = 2 sln2x yields P = 2n'102 where

^ - H * ? - ^ w

a result which can be found In Landau ([6], p. 211). Thus

P = JL . 2 n - l "

This result can also be obtained from the identity

n - l n fi ™Ja JL 2kic\\ 1 1 - cos nt

[1-CM[e + —]\ = ^ l - i - cos e k = i

(given in [2], p. 119) by taking the limit as 9 approaches 0 (using PHopitaPs Rule twice). A related result can be obtained from the identity

n9

n - l

Si ICOS 9 ~ CQSz ^kw) 1 sin n9 J I __ _ _ _ , e i . B .my-

k * l

(given In [1], p. 167, problem 11) by taking the limit as 9 approaches 0 (using one application of PHopitaPs Rule).. This gives

n - l

IK1--*) - £ • k = l

We mention a few related results. Baica has stated in [3] that, for n > 3,

n t1 - ^ r n ' 2

i n-l-k « n - 2 2 ( n - i ) (n-2) / 2

k » l

His remark "Any proof avoiding cyclotomlc fields could be very difficult, if not insoluble11 should represent a challenge to Crux readers. Baica also proved (in [4]) that for n > 2,

n (2 - a Evans [5] has stated that for odd n > 3,

n - 1 fc-i

- 54 -

ti. - 1

J ] (2cos| )k"1 = (-l)C»2-»'8

and

J[ (2cos| ) = (-l)<-'>'2.

References:

[1] J. Heading, Mathematics Problem Book I: Algebra and Complex Numbers, Pergamon Press, New York, 1964.

[2] E.W. Hobson, A Treatise on Plane and Advanced Trigonometry, Dover Publications Inc., New York, 1957.

[3] Malvina Baica, Trigonometric identities, Abstracts of the A.M.S. 7 (1986) 249. [4] Malvina Baica, Trigonometric identities, International Journal of Mathematics

and Mathematical Science 9 (1986) 705-714. [5] R.J. Evans, Review of [4], Math Reviews 88c #11046. [6] E. Landau, Elementary Number Theory, Second Edition, Chelsea, New York,

1966.

Also solved by SBUNG-JIN BANG, Seoul, Republic of Korea; HANS

ENGELHAUPT, Gundelsheim, Federal Republic of Germany; SA'AR HERSONSKY,

student, Technion, Haifa, Israel; RICHARD I. HESS, Rancho Palos Verdes,

California; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; KEE-WAI

LAU, Hong Kong; DAVID G. POOLE, Trent University, Peterborough, Ontario; MA.

SELBY, University of Windsor, EDWARD TH. WANG, Wilfrid Laurier University;

KENNETH M. WILKE, Topeka, Kansas; KENNETH 5. WILLIAMS, Carleton

University; and the proposer. One other reader gave the correct answer without proof

Several readers solved the problem as in solution III by using identity (3). Wang, for instance, referred to exercise 52, p. 25 of Complex Variables in the

Schaum's Outline Series, while Williams gave the reference p. 173 of T. NagelVs

Introduction to Number Theory. Other readers used the identity

m- i n . kw Jrri

among the references cited was problem 143, p. 24 of Yaglom and Yaglom, Challenging Mathematical Problems with Elementary Solutions Vol. II, Dover, New

York, 1967. * * *

- 5.5 -

1397. [1988: 302] Proposed by G.R. Veldkamp, De Bill, The Netherlands. Find the equation of the circle passing through the points other than

the origin which are common to the two conies x2 + 6xy + Wx - 2y = 0 , x2 + Zy2 - 7x + 5y = 0 .

1 Solution by Clayton W. Dodge^ University of Maine, Orono.

The desired circle has the equation 39X2 + 39j/2 - 267x - 59y = 144.

To obtain this circle we derive an equation that has a factor of x and that passes through the four points of intersection of the two conies. Then we divide out the factor of x, removing the point at the origin from this equation, then juggle this equation with the original two equations to get an equation of a circle that passes through the remaining three points common to the two equations. To this end, solve the first given equation for 2y, double the second equation and substitute selectively for 2y} obtaining

2x2 + 3t/(z2 + 6xy + lOx) - Ux + 5(T? + 6xy + 10s) = 0 , which reduces to

x(7x + %xy + 18y2 + 60y + 36) = 0 . Divide by x and then double the resulting equation to get

14x + 6xy + Wy2 + 120y + 72 = 0 . Next subtract the first given equation to eliminate the xy term:

Ax - x2 + Wy2 + 122y + 72 = 0 . Finally double this last equation and subtract the result from 37 times the second given equation to get equal coefficients for the x2 and y2 terms.

II. Solution by the proposer.

We write the equations of the given conies as follows: a(x + 10) + y(6x - 2) = 0 , z(x - 7) + y{3y + 5) = 0 .

The common points other than O are therefore on the conic represented by (x + 10)(3y + 5) - (6* - 2){x - 7) = 0 , .

or 6x2 - Zxy - 49x - 30y - 36 = 0 .

The circle we are looking for has therefore an equation of the form k(x2 + 6xy + lOx - 2y) + m(x2 + $y2 - 7x + 5y)

+ n(6%* - Zxy - 49x - 30y - 36) = 0 . This is actually the equation of a circle if and only if

- 56 -

6k - Zn = 0 and k + m + 6n = 3m , i.e.

n = 2k and 13fc = 2m . Hence km:n = 2:13:4. The equation of the circle is therefore

39z2 + 39j/2— 267s - 59y - 144 = 0 .

;4k0 solved by HAYO AHLBURG, Benidorm, Spain; FRANCISCO BELLOT

ROSADO, LB. Emilio Ferrari, VaUadolid, Spain; HANS ENGELHAUPT,

Gundelsheim, Federal Republic of Germany; RICHARD I. HESS, Rancho Palos

Verdes, California; KEE-WAI LAU} Hong Kong; and P. PENNING, Delft, The Netherlands. Two incorrect solutions were sent in.

* * *

1398. [1988: 302] Proposed by Ravi Vakil, student, University of Toronto.

Let p be an odd prime. Show that there are at most (p + l)/4 consecutive quadratic residues mod p. For which p is this bound attained?

Solution by Kee-Wai Lau, Hong Kong.

We need the following well-known results. I) There are

p - 4 - ( l ) (P- l )^2

numbers a such that both a and a + 1 are quadratic residues mod p. II) There are

p - 2 + (-!)"»•»'» 4

numbers b such that both 6 and b + 1 are quadratic nonresidues mod p.

III) For p = 4m + 3, g is a quadratic residue mod p if and only if p - g

is a quadratic nonresidue mod p.

IV) i) The product of two quadratic residues or nonresidues is a

residue mod p.

ii) The product of a quadratic residue and a quadratic nonresidue is a quadratic nonresidue mod p.

Hence by I there are at most (p - l)/4 consecutive quadratic residues if p = 4m + 1, and at most (p + l)/4 consecutive quadratic residues if p = 4m + 3.

We now proceed to find those primes p = 4m + 3 for which the bound (p + l)/4 is attained. By direct calculations we find that the bound is attained for p = 3, 7 and 11, but not attained for p = 19. From now on we assume that p > 19, i.e. m > 5.

- 57 -

Suppose that the bound is attained for Home such p = 4m + 3. Let S = {1,2,-•-,2TH + 1}; T = {2m + 2,2m + 3}---y4m + 2}? R the set of (p + l)/4 = m + 1 consecutive quadratic residues and (by III) N the corresponding set of m + 1 consecutive quadratic nonresidues. By 1 and II there are no other consecutive residues or nonresidues. Hence by III either (a) R c Sf N c T or (b) N C 5, fl C T. Denote residues by r and nonresidues by n.

Case (a). If the pattern

1 2 3 ••• m+ 1 m+ 2 m+ 3 m + 4 m+ 5 m + 6 •••

r r r . . . r • 7i r fi r fi

holds? we deduce from IV (i) (since m > 2) that m + 2? m + 4 and m + 6 are prime numbers, which is impossible. Otherwise? the pattern

1 2 ••• i - 2 $-1 f J + l i + 2 4+3 •-• £+m t+m+1 r n r n T r r T r n

holds for some odd £? 3 < t < m + 1. Let 5 > 0 be the greatest integer such that s(s + 1) < t Then

* < (5 + l)(s + 2) < « + m , since

(s + l)(s + 2) = s(5 + 1) + 2(5 + 1) < t + 2(5 + 1) I t + 5 ( 5 + 1 ) <2t~l<i+m if 5 > 2 ,

l ~ | i + 4 < i + m if 5 = 0 or 1 .

Also t if 5 = 0 or 1 j

s($ + 1) < t if 5 > 2 . Since (5 + 1)(5 + 2) is a quadratic nonresidue by IV (ii) we arrive at a contradiction.

Case (b). We have the pattern 1 2 3 4 -•- t - 1 . t t + 1 --• t + m r T i r n r n n n

for some even J, 2 < t < m + 1. If J = 2, then 6 = 2-3 is a residue by IV (i), which is a contradiction since m > 4. If t > 43 then t + 1, i + 3 and i + 5 are odd and (since m > 5) nonresidues. By IV (i) they must be prime numbers, which is a contradiction.

Thus equality is attained only for p = 3, 7 and 11.

5 + 2 <

- 58 -

Also solved by WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria]

and by PAT SURRY, student, University of Western Ontario, and the proposer. The

first part only was solved by KENNETH S. WILLIAMS, Carleton University.

The fact I) in Lau9s solution was mentioned by both Janous and Williams, the

latter giving the reference LM. Vinogradov, Elements of Number Theory, p. 97. It is known (D.A. Burgess, The distribution of quadratic residues and non-

residues, Mathematika 4(1957) 106-112) that the maximum number R(p) of

consecutive quadratic residues modulo the prime p satisfies

R(p) < p1M*& times a constant

for any 6 > 0. For more information on the problem of estimating R(p), see p. 136 of R.K. Guy, Unsolved Problems in Number Theory, Springer-Verlag, 1981.

1399. [1988: 302] Proposed by Sydney Bulman-Fleming and Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. Prove that

W < ia+M for all natural numbers n and determine all cases when equality holds. (Here a(k)

denotes the sum of all positive divisors of k.)

Solution by Robert E. Shafer, Berkeley, California.

It can be shown directly that the inequality holds for n < 7, with equality for n - 1,2,3,4,5. Thus assume n > 8. Let

n! = 2a i3a25a3-..p , where p is the largest prime less than or equal to n. Then

,2 _ 1 a(n\) = — r — 2 — "

= n\ (0ai+l _ , ofl2+l _ i

2ai 3 5 Ti

2-3 C2

P ~ 1

P2 - 1 [P ~ 1)P

.(2 3 5 p \ (i)

(2 3 4 5

\rrrr n_

' n -4 6 8 9

T n +

n 1\

/

< n!

(n + 1)! . (n + 1)! " 72/35 < 2~~^ '

Thus the equality holds for 1 < n < 5 and for no other values of n.

- 59 -

The reader may recall the problem of estimating the product

IIM) in the computation of

2<p<n

Also solved by SA'AR HERSONSKY, student, Technion, Haifa, Israel;

RICHARD I. HESS, Rancho Palos Verdes, California; KEE-WAI LAU, Hong Kong; C WILDHAGEN, Breda, The Netherlands; and the proposers. Two other readers sent in short induction arguments that the editor found suspect

The proposers' proof was similar to Shafefs (but not quite as elegant). They

remark that a much sharper upper bound for a(nl) is known. Namely, there exists a

constant c > 0 such that for all n > 2,

(e.g. Theorem 9.1A.1, p. 324 of H.N. Shapiro, Introduction to the Theory of Numbers), and hence (from, (1) in Shafefs proof)

a ( n ! ) < zLi2S_n . c

They say, however, that the proof does not seem to be easy. The problem was

motivated by the easy lower bound a(nl) > nlfl + \ + ••• + ~]

(e.g. problem 10(b) in Chapter 6 of D.M. Burton, Elementary Number Theory). * * *

1400. [1988: 302; 1989: 177] Proposed by Robert E. Shafer, Berkeley,

California.

In a recent issue of the American Mathematical Monthly (June-July 1988, page 551), G. Klambauer showed that if xse"x = yse"y (x, y, s > 0 , x $ y)

then x + y > 2s. Show that if xse"x = yse"y where x # y and x, y, s > 0 then xy(x + y) < 2a8.

I. Solution by the proposer (slightly modified by the editor).

Suppose on the contrary that

xy(x + y) = 2Xs2

where A > 1. Assuming without loss of generality that x > y, we put

- 60 -

yx2 = As3(l + t)

xy2 = As3(l - t)

where 0 < t < 1. Then / \ x2y (xy = \xW = a + ty = eSiog[(iu),(i-t)])

while

Thus

x_y=yJ-xy2 = y xy

\y2x)

xy = y yJ-xy2 = s2>$ A2(l - t2) .

2XsH n_± I T = 2s*-

s2-J A 2 ( l - i 2 ) 3 f l - * 2

Now recall from the proof of Crux 1247 [1988: 191] that for 0 < t < 1,

(l - ^ ) ^ 3 l1 - ^

from which we find (since A > 1)

f^]S = 6 S l O g [ ( l U ) x ( l - t ) ]

< g2st^( 1-t )

« w , , 2 i 1 ' 3

< c 2 « t [ \ x ( l - t ) ]

or xse"x < yse'y} a contradiction.

II. Solution by Murray S. Klamkin, University of Alberta.

Note that if xse'z= yse~y then s " In x - In y '

the so-called logarithmic mean of x and y, and that the Klambauer result alluded to goes back at least to Ostle and Terwilliger [3]. A stronger inequality

s < T

is due to Lin [2]. For an inequality going the other way, Carlson [1] has shown that

Unfortunately, the latter inequality is not comparable with the proposed problem

here.

- 61 -

The log mean of x and y is not defined for x = y but its definition can be extended to be x for this case. Then this function would be continuous for all x, y > 0.

The proposed inequality reduces to the form

2\x - y\3 > xy{x + y)\ln\x/y)\ , x, y > 0 , with equality if and only if x = y. We can assume without loss of generality that x > y and replace x/y by e* where t > 0 to give the equivalent inequality

F(t) = 2(e* - I)3 - *V(e* + 1) > 0 . (1) Since F(0) = 0, one standard method of establishing inequalities such as (1) is

to show that F'(t) > 0. Since the latter is not obvious and F'(0) = 0, we try to show F"(t) > 0. If necessary, we continue differentiating (it turns out we have to differentiate quite a number of times). Along the way we can discard any non-negative factors for simplification. Here,

F'(t) = 6(e* - 1 )V - 3*2(e* + e2t) - *3(e* + 2e2t) .

We can discard the factor 6* and then consider G{t) = 6(e* - I)2 - 3i2(l + e*) - f3(l + 2e*).

Here G(0) = 0 and H(t) = G'^e* = 12(g* - 1) - [6t + 9<2 + 2t3 + (6* + 3*2)e_t].

Here H(0) = 0 and H'(t)jZ .= 4e* - [2 + 6t + 2f2 + (2 - ^e"*] .

Here ff'(0) = 0 and # ' (0 /3 = 4e* - [6 + it + (*2 - 2t - 2)e-t] .

Here H"(0) = 0 and

JT(*)/3 = 4«* - 4 + (i2 - 4<)e"* > 0 . Here #"(0) = 0 and finally

H"{t)lZ = e-*(4e2t - *2 + 6< - 4) > 0 by expanding out e2t. Hence F(t) is an increasing function, so F(t) > 0 for t > 0.

Another way of proceeding is to expand out (1) and replace e3t, e2t and e* by their power series. Doing this we end up with the power series

42£ + 504^ + ^ + aioti0 + . . . f

where for n > 2

nla* = 2-3n - 6-2n + 6 - n(n - l)(n - 2)(2ft-3 + 1) .

To show that cn > 0 for n > 8, it suffices to show that

* . 2 - [6 + «(" - iy» - 2)](§)° - "(" - 1)(" - ») > 0 .

- 62 -

But this follows inductively from

* „ - ,. . [j + "(»-l|(tt-8)](|" + «(• - j ) ( * - *) > 0

for n > 8.

.Re/erences: [1] B.C. Carlson, The logarithmic mean, Amer. Math. Monthly 79 (1972) 615-618. [2] T.P. Lin? The power mean and the logarithmic mean, Amer. Math. Monthly 81

(1974) 879-883. [3] B. Ostle and ILL. Terwilliger, A comparison of two means, Proc. Montana

Acad. Sci. 17 (1957) 69-70.

Also solved by RICHARD I. HESS, Rancho Palos Verdes, California; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; and KEE-WAI LAU}

Hong Kong. There was one incorrect solution sent in.

Janous notes that Klambauefs and the proposefs inequalities combined have the curious interpretation

aj *.y.^-K < - < l^+V + ^V1) or

M0[x,y,^^ <s < tfi(*,y,^),

where as usual

(ul + vl + wl\u l

I T "J t f o {uvw)l/l , t = 0

is the t-th mean of the positive numbers u, v, w. * * *

1 4 0 1 [1989: 12] Proposed by P. Penning, Delft, The Netherlands.

Given are a circle C and two straight lines I and m in the plane of C

that intersect in a point S inside C. Find the tangent(s) to C intersecting I and m

in points P and Q so that the perimeter of ASPQ is a minimum.

Solution by Jordi Dou, Barcelona, Spain. Let Ct, C2, C%, Ci be the circles tangent to I and m, and externally tangent

to C. Let Mi be the point of contact of Ci with m, for i = 1,2,3,4, and let SMy.

Mt{u,v,w) =

- 63 -

be the smallest of the segments SM{. The tangent t which we look for is the tangent common to circles C and C* at their point of contact.

Notice that C\ is an excircle of the triangle enclosed by lines Z, TO, t, and hence 2SM& is the perimeter of this triangle. If V i t is a tangent to C which cuts the half-lines SI and 5m, then the excircle C\ (opposite S) of the triangle enclosed by f, ms

tf will be exterior to and not tangent to C, and therefore its point of contact M\ with m will be such that SM^ > SM^.

The construction of circles C% is known and classical.

Also solved by HANS BNGELHAUPT, Franz-Ludwig-Gymnasium, Bamberg,

Federal Republic of Germany] MARGIN E. KUCZMA, Warszawa, Poland; and the

proposer. All proofs were similar,

Kuczma's solution included a method (unfortunately algebraic) for determining

which of the four circles tangent to C yields the correct tangent line. * * *

1402. [1989: 12] Proposed by George Tsintsifas, Thessaloniki, Greece. Let M be an interior point of the triangle AtA2A^ and Bh B2i B% the

feet of the perpendiculars from M to sides A2A^ A$Ah AtA2 respectively. Put r» = BiM, i = 1,2,3. R' is the circumradius of ABtB2Bh and R} r the circumradius and inradius of AA\A2A%. Prove that

R'Rr > 2r\r2r% •

I. Solution by Walther Janous, Ursulinengymnasium, Innsbruck, Austria.

In [2], p. 344, 1.4, 6) the following stronger result of Carlitz is given:

2R'R2 > R\R2Ri , (1) where Ri = AiM, i = 1,2,3. To see that (1) is better we refer to item 12.26 of [1],

RlR2R, > W l f ( 2 )

where ^ ^

s - nsin = & • (3)

- 64 -

Then (l)-(3) yield 2R)R2 y 4fir1r2r3

r ' i.e. R'Rr > 2r1r2r3. Done.

It should be noted that (1) is a disguised way of writing F > 4F' ,

where F and F' are the areas of triangles AxA^A^ and 818283, respectively (item 9.5 of [1]).

II. Solution by Marcin E. Kuczma, Warszawa, Poland.

This is misleading! The inequality that actually holds is

R12 > 2nr2n . (4) The given inequality hence results automatically, as (trivially) R' > r. Inequality (4) is an immediate consequence of

2s < RJT! (s the semiperimeter of AA1A2A3) and

2F2

rxr2rz < ^

(F - rs the area of AAiA2As), which are respectively items 5.3 and 12.29 of [1]. Equality holds in (4) only in the case that AA\A2Az is equilateral and M is its center.

References:

[1] 0. Bottema et al, Geometric Inequalities.

[2] D.S. Mitrinovic, J.E. Pecaric and V. Volenec, Recent Advances in Geometric Inequalities, Dordrecht, 1989.

Also solved by FRANCISCO BELLOT ROSADO, LB. Emilio Ferrari,

Valladolid, Spain; SVETOSLAV J. BILCHEV and EMILIA A. VELIKOVA, Technical

University, Russe, Bulgaria] MURRAY S. KLAMKIN, University of Alberta] and the

proposer.

Klamkin gave the same improvement (1) to the problem as Janous did, only in

the form

ata2az > 4 V a\r2r% -

He noted that it appeared earlier in Crux [1987: 260]. * * *

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